Discussion Forum

On a photosensitive material, when the frequency of incident radiation is increased by 30%, kinetic energy of emitted photoelectrons increases from  0.4 eV to 0.9 eV. The work function of the surface is 

Answer:

Explanation:

 According to the Einstein equation,

  $KE_{max}=hv_{0}-\phi _{0}$

 where,  $KE_{max}$   = maximu kinetic energy and $\phi_{0}$ 

                                               = work function

 Initially , hv= 0.4+ $\phi _{0}$   ........(i)

When the frequency of incident radiation  is increased  by 30%  , then 

     1.3 hv=0.9 +$\phi _{0}$          .........(ii)

 Solving Eqs.(i) and (ii) , we get

$0.3\phi _{0}=0.9-1.3(0.4)$

$\therefore$      $\phi _{0}=\frac{0.38}{0.3}=1.267 eV$